And just for future reference for anyone who stumbles across a similar problem, I ended up with:
P = P_{0} + \frac{1}{\gamma{V}} + \frac{\beta}{\gamma{V}}e^{-C_{v}}e^{-\frac{\beta}{\gamma}(V-V_{0})}
dP/dV of that should give the slope of the adiabat!
So solving dS gives me:
dS = \frac{dT}{T-T_{0}} + \frac{\beta}{\gamma} = 0
and subbing into dV gives:
-\frac{\gamma}{\beta}\frac{dT}{T-T_0} = \frac{\partial{V}}{\partial{T}}dT + \frac{\partial{V}}{\partial{P}}dP
But from here, and I'm probably being thick, I find myself with a lot of...
Probably should have included that in my explanation. I get that for an adiabatic change, dS is 0 assuming reversibility. Similarly, I got to the point where I have:
dV = \frac{\partial{V}}{\partial{T}}dT + \frac{\partial{V}}{\partial{P}}dP
and
dS = \frac{\partial{S}}{\partial{T}}dT +...
Homework Statement
The Equation of State and the expression for the entropy for a sample of salt water is given by:
V = V_{0}(1 + \beta(T - T_{0}) - \gamma(P - P_{0})) S = S_{0} + C_{v}ln(T - T_{0}) + \frac{\beta}{\gamma}(V - V_{0})
where the subscript 0 denotes a reference state, the...